Problem: $\overline{AC}$ is $12$ units long $\overline{BC}$ is $5$ units long $\overline{AB}$ is $13$ units long What is $\sec(\angle BAC)?$ $A$ $C$ $B$ $12$ $5$ $13$
Answer: $\sec(\angle BAC) = \dfrac{1}{\cos(\angle BAC)}$ How can we find $\cos(\angle BAC)$ SOH CAH TOA osine = djacent over ypotenuse Adjacent $= \overline{AC} = 12$ Hypotenuse $= \overline{AB} = 13$ $\cos(\angle BAC) = \dfrac{12}{13}$ $\sec(\angle BAC) = \dfrac{1}{\cos(\angle BAC)} = \dfrac{13}{12}$